A stone is thrown verticaly upward with an initial velocity of `40m//s`. Taking `g=10m//s^(2)`, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

A stone of 3kg is thrown vertically upward with an initial speed of 20m/s. If g=10 m//s^(2) , then find the kinetic energy and potential energy after 1s.

A stone is thrown vertically upwards with a velocity of 4.9 ms^(-1) . Calculate : the time taken to reach the maximum height.

A stone is thrown vertically upwards with a velocity of 4.9 ms^(-1) . Calculate : the maximum height reached.

A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone. (b)Find its velocity one second before it reaches the maximum height. (c)Does the answer of part b change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s?

A stone thrown vertically upwards rises S ft in t seconds where S =112t-16t ^(2) . The maximum height reached by the stone is

A stone is thrown vertically upwards with an initial velocity u from the top of a tower of lieight (12u^(2))/(g) With what velocity does the stone reach the ground?

A particle is thrown vertically upward. Its velocity at half of the height is 10 m/s. Then the maximum height attained by it : - (g=10 m//s^2)

A stone is projected vertically upwards.When it reaches half the maximum height,its velocity is 20m/s.What is the maximum height attained by the stone? (g=10m/s^(2)